I = ∫ 8x/4cos^2x+sin^2x dx x ∈ to (0, π) equals to

Question

\(I=\int_\limits{0}^{\pi} \frac{8 x}{4 \cos ^{2} x+\sin ^{2} x} d x\) equals to

(1) \(\pi^{2}\)

(2) \(4 \pi^{2}\)

(3) \(2 \pi^{2}\)

(4) \(\frac{3}{2} \pi^{2}\)

Answer 1

Correct option is: (3) \(2 \pi^{2}\)  

Put tanx = t 

\(\sec ^{2} x d x=d t \)

\(I=8 \pi \int_\limits{0}^{\infty} \frac{d t}{4+t^{2}} \)

\(I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_{0}^{\infty} \)

\(I=4 \pi\left(\frac{\pi}{2}\right) \)

\(I=2 \pi^{2}\)